Answer:
The ionization energy (in kJ/mol) of the helium ion is 21,004.73 kJ/mol .
Step-by-step explanation:
![E_n = -(2.18 10-18 J)* (Z^2)/(n^2)](https://img.qammunity.org/2020/formulas/chemistry/high-school/ye16tp63ma8az035s7xfx9el35rucj69nu.png)
Z = atomic mass
n = principal quantum number
Energy of the electron in n=1,
![E_1= -(2.18 10^(-18) J)* (4^2)/(1^2)=-3.488* 10^(-17) J](https://img.qammunity.org/2020/formulas/chemistry/high-school/rmxsajeyyjm82l8zpddokva1cigcakgikp.png)
Energy of the electron in n = ∞
![E_(\infty)= -(2.18 10^(-18) J)* (2^2)/(\infty ^2)=0 J](https://img.qammunity.org/2020/formulas/chemistry/high-school/f0oxnlcdkksdgea3vs6elyraonb0kacnxz.png)
Ionization energy of the
ion:
![I.E=E_(infty)-E_1=0-(-3.488* 10^(-17) J)=3.488* 10^(-17) J](https://img.qammunity.org/2020/formulas/chemistry/high-school/681fzee01na6gqgao6e99rm4xdd5az6so1.png)
![I.E=3.488* 10^(-20) kJ](https://img.qammunity.org/2020/formulas/chemistry/high-school/5ql0v9a4jcb5nlxk5np2w1w6m6gdbp3emq.png)
To convert in into kj/mol multiply it with
![N_A=6.022* 10^(23) mol^(-1)](https://img.qammunity.org/2020/formulas/chemistry/middle-school/jpzydvgwzrkshnotcheiqbkxnt1kaaf2ms.png)
![I.E=3.488* 10^(-20) kJ* 6.022* 10^(23) mol^(-1)=21,004.73kJ/mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/t6rm0nmkkzs4repzfrs6q42t1tepxvnvbp.png)